Question

FOR WHAT VALUE OF K , THE NUMBER 3K+2, 4K+3, 6K-1 ARE IN AP.

Answer 1

3k+2,4k+3,6k-1 are in arithmetic progression
a2 - a1 = a3-a2
4k+3-(3k+2)=6k-1-(4k+3)
4k+3-3k-2=6k-1-4k-3
k+1=2k-4
1+4=2k-k
5=k
Therefore
k=5

Answer 2

Answer:

3k+2,4k+3,6k-1

a2 - a1 = a3-a2

4k+3-(3k+2)=6k-1-(4k+3)

4k+3-3k-2=6k-1-4k-3

k+1=2k-4

1+4=2k-k

5=k

k=5